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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2009 HOMEWORK ASSIGNMENT 7: 1. Spontaneous emission: In this problem, we will use the Fermi Golden Rule to estimate the spontaneous emission rate of an atom. We consider an atom which is initially excited, and therefore has energy E e . It decays to the ground state, having E g = 0, by emitting a photon of energy planckover2pi1 . To use the FGR, = 2 planckover2pi1  G ( E i )  2 n ( E i ) , we need the density of photon states, n ( planckover2pi1 ), as well as the atomfield coupling constant, G ( planckover2pi1 ). Clearly the decaying atom cannot tell whether it lives in an infinite space, or in a very large box. Hence the decay rate should be the same in either situation, and should be independent of L , the box length. We know from mathematics that any function defined on the interval [0 ,L ] which vanishes at the edge points, can be expanded in terms of sin( k n x ) where k n = n/L , with n being any positive integer. If the box is large enough, then the photon electric field must vanish at the walls. This field can be therefore be expanded onto the set of modes, u n x ,n y ,n z ( x,y,z ) = sin( n x x/L )sin( n y y/L )sin( n z z/L ) , where the energy of each photon mode is E = planckover2pi1 ck = planckover2pi1 c ( L ) radicalBig n 2 x + n 2 y + n 2 z . a.) To determine the density of states at energy E , first determine N ( E ), which is clearly the number of ( n x ,n y ,n z ) points with energies below E . The density of states can then be computed via n ( E ) = dN ( E ) dE . Answer: From the box eigenmodes, we see that n x , n y , and n z are positive integers. Since k n = n/L , we see that the spacing, k , is given by k = /L . Thus with vector k n x n y n z = L ( n x vector e x + n y vector e y + n z vectore z ) we see that the allowed kvectors lie on a 3d lattice, with spacing /L . The question is then how many lattice points are inside a sphere of radius k = E planckover2pi1 c , subject to the constraint { k x ,k y ,k z } > 0? So the density of points is = 1 point ( /L ) 3 = L 3 3 The volume of the energy sphere is V ( E ) = 4 3 k 3 ( E ) so the number of points satisfying the two conditions is N ( E ) = 1 8 V ( E ) = 1 8 4 L 3 k 3 ( E ) 3 2 = L 3 E 3 6 2 planckover2pi1 3 c 3 1 This leads to a density of states of n ( E ) = d dE N ( E ) = L 3 E 2 2 2 planckover2pi1 3 c 3 2 b.) To determine the coupling constant, we will start from the electric dipole energy E = vector d vector E , where E is the electric field amplitude and d is the atomic dipole moment. For the atomic dipole moment we can use d = ea , while for E , we need to find the electric field of a single photon....
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This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.
 Fall '08
 MichaelMoore
 mechanics, Work

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